8 - L. Karp Notes for Dynamics VIII. "Nonconvex"...

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L. Karp Notes for Dynamics VIII. "Nonconvex" Control Problems 1) Describe growth model that leads to non-convex control problem 2) Sketch phase portrait of solutions to FOCs 3) Identify optimal candidate. 4) Economic interpretation. 5) A pollution control problem with similar features. 6) Sketch phase portrait 6) A different way to identify optimal trajectory. In this section we study a control problem with increasing returns to scale (IRTS) of the state equation over some interval. For some initial conditions it is optimal to drive the state to 0 (e.g., wipe out the resource) and for some initial conditions it is optimal to build up the state. In this problem typicaly 5 multiple solutions to FOC’s, and SOC’s do not hold. How do you identify the optimal solution? 1) Growth model with IRTS (Brock and Malliaris, pp 159 - 168) 2 sector optimal growth model x = capital stock, x ˙ = investment x 1 = amount of capital in "neoclassical" sector (concave production function) y 1 = production of power (an intermediate good) g 1 ( x 1 , y 1 ) = output of capital / consumption good, a neoclassical prod’n function y 1 = g 2 ( x - x 1 ) increasing and convex. Economy can allocate capital between sectors U ( g 1 ( x 1 , y 1 )- x ˙ ) = utility objective: max 0 e ρ t U g 1 ( x 1 , y 1 ) ˙ xd t Define (1) g ( x ) max x 1 g 1 x 1 , g 2 ( x x 1 ) This is maximum output of final good, given x . Assume g ( x ) convex-concave (M & B show this is the case).
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Rewrite problem as max 0 e ρ t U ( c ) dt c + x ˙ = g ( x ) x 0 given We will show that if initial level of capital is low, it is optimal to run it down (stagnate). If initial level is sufficiently high, it is optimal to build it up (save). We begin by looking at the characteristics of the function g(x) Figure 1 defined as lowest value of x where ρ ( x )= g ( x ) x ~ FOC to (1): x 0 1 solves (1) g 1 x 1 g 1 y 1 g 2 0 envelope thrm g g 1 y 1 g 2 g 1 x 1 8:2
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Figure 2 AP g ( x ) x MP g ( x ) Now we begin to look at the control problem. Here are the Hamiltonian and necessary conditions. (2) (current value) H U ( c ) λ g ( x ) c (3) U ( c ) λ⇒ c C ( λ ) C 1 U ′′ <0 (4) ˙ λ λ ρ g ( x ) (5) ˙x g ( x ) C ( λ ) 8:3
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2. Sketch phase portrait We will sketch the portrait in x, λ space. First we find the isocline(s) for λ ˙ =0 . λ ˙ ⇒ρ = g ( x ) x = x or x = x . for x < x < x , g ( x )> ρ⇒λ ˙ < 0. These results are summarized in figure 3. Figure 3 Now we find the isocline for x ˙=0 . x g ( x )- C ( λ )=0 g ( x ) dx C d λ ( < 0 when g ( x ) > 0 since C < 0; see eqn 3) d λ dx ˙x 0 g ( x ) C So we know that the isocline for x ˙ = 0 is downward sloping, as shown in figure 4. Now we have find the directional arrows; i.e., is x increasing or decreasing above the isocline? To determine
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8 - L. Karp Notes for Dynamics VIII. &quot;Nonconvex&quot;...

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